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Max Alekseyev
Max Alekseyev
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It needs to be pointed out that the series $\frac{\sin(x)-x\cos(x)}{x^3}$ comes from the MSE answer, and as I understand the claim there this series equals $\prod_{k\geq 1} (x^2-\lambda_k^2)$$\frac{1}{3}\prod_{k\geq 1} (1-\frac{x^2}{\lambda_k^2})$. I have not verified this claim, but if it's true, then your approach does make sense. UPD. Precise argument can be seen in this answer.

I've also verified your analytical expression numerically, and it looks fine. Namely, the analytical expression gives $\sum_k a_k^{-2} = 0.0973745978\ldots$, while numerically we have $\sum_{k=1}^{10^5} \frac {\lambda _k ^2 +1}{\lambda _k ^2 (\lambda _k ^2 +2)} = 0.09737358469\ldots$ with the difference $\approx 10^{-6}$.

It needs to be pointed out that the series $\frac{\sin(x)-x\cos(x)}{x^3}$ comes from the MSE answer, and as I understand the claim there this series equals $\prod_{k\geq 1} (x^2-\lambda_k^2)$. I have not verified this claim, but if it's true, then your approach does make sense.

I've also verified your analytical expression numerically, and it looks fine. Namely, the analytical expression gives $\sum_k a_k^{-2} = 0.0973745978\ldots$, while numerically we have $\sum_{k=1}^{10^5} \frac {\lambda _k ^2 +1}{\lambda _k ^2 (\lambda _k ^2 +2)} = 0.09737358469\ldots$ with the difference $\approx 10^{-6}$.

It needs to be pointed out that the series $\frac{\sin(x)-x\cos(x)}{x^3}$ comes from the MSE answer, and as I understand the claim there this series equals $\frac{1}{3}\prod_{k\geq 1} (1-\frac{x^2}{\lambda_k^2})$. I have not verified this claim, but if it's true, then your approach does make sense. UPD. Precise argument can be seen in this answer.

I've also verified your analytical expression numerically, and it looks fine. Namely, the analytical expression gives $\sum_k a_k^{-2} = 0.0973745978\ldots$, while numerically we have $\sum_{k=1}^{10^5} \frac {\lambda _k ^2 +1}{\lambda _k ^2 (\lambda _k ^2 +2)} = 0.09737358469\ldots$ with the difference $\approx 10^{-6}$.

Post Undeleted by Max Alekseyev
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Max Alekseyev
Max Alekseyev
  • 34.3k
  • 5
  • 74
  • 152

It needs to be pointed out that the series $\frac{\sin(x)-x\cos(x)}{x^3}$ comes from the MSE answer, and itas I understand the claim there this series equals $\prod_{k\geq 1} (x^2-\lambda_k^2)$. SoI have not verified this claim, thebut if it's true, then your approach you used does make sense.

I've also verified your answeranalytical expression numerically, and it looks fine as well. Namely, the analytical expression gives $\sum_k a_k^{-2} = 0.0973745978\ldots$, while numerically we have $\sum_{k=1}^{10^5} \frac {\lambda _k ^2 +1}{\lambda _k ^2 (\lambda _k ^2 +2)} = 0.09737358469\ldots$ with the difference $\approx 10^{-6}$.

It needs to be pointed out that the series $\frac{\sin(x)-x\cos(x)}{x^3}$ comes from the MSE answer and it equals $\prod_{k\geq 1} (x^2-\lambda_k^2)$. So, the approach you used does make sense.

I've also verified your answer numerically, and it looks fine as well. Namely, the analytical expression gives $\sum_k a_k^{-2} = 0.0973745978\ldots$, while $\sum_{k=1}^{10^5} \frac {\lambda _k ^2 +1}{\lambda _k ^2 (\lambda _k ^2 +2)} = 0.09737358469\ldots$ with the difference $\approx 10^{-6}$.

It needs to be pointed out that the series $\frac{\sin(x)-x\cos(x)}{x^3}$ comes from the MSE answer, and as I understand the claim there this series equals $\prod_{k\geq 1} (x^2-\lambda_k^2)$. I have not verified this claim, but if it's true, then your approach does make sense.

I've also verified your analytical expression numerically, and it looks fine. Namely, the analytical expression gives $\sum_k a_k^{-2} = 0.0973745978\ldots$, while numerically we have $\sum_{k=1}^{10^5} \frac {\lambda _k ^2 +1}{\lambda _k ^2 (\lambda _k ^2 +2)} = 0.09737358469\ldots$ with the difference $\approx 10^{-6}$.

Post Deleted by Max Alekseyev
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Max Alekseyev
Max Alekseyev
  • 34.3k
  • 5
  • 74
  • 152

It needs to be pointed out that the series $\frac{\sin(x)-x\cos(x)}{x^3}$ comes from the MSE answer and it equals $\prod_{k\geq 1} (x^2-\lambda_k^2)$. So, the approach you used does make sense.

I've also verified your answer numerically, and it looks fine as well. Namely, the analytical expression gives $\sum_k a_k^{-2} = 0.0973745978\ldots$, while $\sum_{k=1}^{10^5} \frac {\lambda _k ^2 +1}{\lambda _k ^2 (\lambda _k ^2 +2)} = 0.09737358469\ldots$ with the difference $\approx 10^{-6}$.